Radius of Curvature and Rectification

Home , Radius of curvature

Study material of B.Sc.(Semester - I) US01CMTH02 (Radius of Curvature and Rectiﬁcation) Prepared by Nilesh Y. Patel Head,Mathematics Department,V.P.and R.P.T.P.Science College

US01CMTH02 UNIT-2

1. Curvature Let f : I → R be a sufficiently many times differentiable function on an interval I. Then the points on the graph of y = f(x) is a curve. However, not all curves could be represented as a graph of such a real valued function on intervals viz, the figure of a circle with centre (0,0) and radius 1 in the XY -plane R2 is one such example of a curve. In this situation, we have to represent the equation of the circle as x = cos t; y = sin t, t ∈ [0, 2π]. These are called the parametric equations of the circle. Also, let us think of a spring put in R3. Then the points of this spring is a curve. Thus formally we have the following definition of a curve.

1.1. Definition. Let I be a closed interval and x = x(t), y = y(t) and z = z(t) be real- valued differentiable functions defined on I. Then the points (x(t), y(t), z(t)) in the space is called a locus of the curve represented by the parametric equations x = x(t), y = y(t), z = z(t), t ∈ I.

Throughout this chapter we shall be concerned only with curves lying in the XY - plane. For such curves we have z = 0. Hence they are described by x = x(t) and y = y(t). A curve lying only in one plane is called a planer curve.

1.2. Deﬁnition. Let x = x(t), y = y(t) be a curve. If we eliminate t and obtain a relation g(x, y) = 0, then this form is called the cartesian representation of the curve. Further, if g(x, y) = 0 can be written in the form y = f(x) (respectively, x = f(y)), then y = f(x) (respectively, x = f(y)) is called the cartesian equation of the curve.

1.3. Example. Let I = [0, 1] and x = t, y = t2. Then this is a curve that can also be represented by the cartesian equation y = x2.

1.4. Deﬁnition. Let y = f(x) be a curve. Fix a point A on this curve. For a point P on the curve, let s = arc AP be the arc length from A to P . For a point Q on the curve, 1 2

let s + ∆s = arc AQ so that ∆s = arc PQ. Let ℓ1, ℓ2 be the tangents to the curve at the points P and Q making angles ψ and ψ + ∆ψ respectively, with

a ﬁxed line in the plane. Y ↑ Clearly, the angle between these two tangents is ∆ψ, called the total bending or total curvature of the arc be- ℓ2 tween P and Q. Hence the ℓ1 Q ∆ψ average bending or the av- ∆s erage curvature of the curve s A P between these two points ← ψ ψ + ∆ψ → relative to the arc length is O↓ X ∆ψ given by ∆s . The bending or Figure 1.4 the curvature of the curve at P is deﬁned to be dψ = lim ∆ψ = lim ∆ψ . ds Q→P ∆s ∆s→0 ∆s 2. Derivative of an arc

2.1. Proposition. Fix a point A(x0, y0) on a curve given by y = f(x). For a point P (x, f(x)) on the curve, let s be the arc length of arc AP . (Clearly, s is a function of x.) Then prove that √ ( ) ds dy 2 = 1 + . dx dx 2. Derivative of an arc 3

Proof. Let y = f(x) represent the Y ↑ Q given curve and A be a ﬁxed point on it. Let P (x, y) be a generic point on the curve. ∆y Let the arc AP = s. Take A ∆s s P a point Q(x + ∆x, y + ∆y) N on the curve near to P . ∆x Let arc AQ = s + ∆s. From the right angled triangle △PNQ, we have, ← → O↓ L M X Figure 2.1 PQ2 = PN 2 + NQ2 = (∆x)2 + (∆y)2 ( ) ( ) PQ 2 ∆y 2 ⇒ = 1 + ∆x ∆x ( ) ( ) ( ) chord PQ 2 ∆s 2 ∆y 2 ⇒ = 1 + . arc PQ ∆x ∆x Taking Q → P , we get chord PQ → arc PQ. Hence, ( ) ( ) √ ( ) ds 2 dy 2 ds dy 2 = 1 + ⇒ = 1 + . dx dx dx dx

The proof of the following corollary is left to the reader.

2.2. Corollary. Let x = x(t) and y = y(t) be the parametric equations of a curve. Then √( ) ( ) ds dx 2 dy 2 = + . dt dt dt

2.3. Exercise. In the Proposition 2.1, suppose that the curve is represented by x = f(y). √ ( ) 2 ds dx Then deduce that the derivative of the arc length dy = 1 + dy .

2.4. Deﬁnition. Let h(x, y) = 0 be a cartesian representation of a curve. By substituting x = r cos θ and y = r sin θ, in this form we get a representation g(r, θ) = 0 of the curve called a polar representation of the curve.

We shall be mainly dealing with the form r = f(θ) of the curve.

2.5. Theorem. For a polar equation r = f(θ) of a curve, √ ( ) ds dr 2 = r2 + . dθ dθ 4

Proof. Let r = f(θ) represent the given curve and A be a ﬁxed point on it.

Let P (r, θ) be a generic Y ↑ point on the curve. Let Q(r + ∆r, θ + ∆θ) N s arc AP = s. Take a ∆ point Q(r + ∆r, θ + ∆θ) r P (r, θ) on the curve near to P . + ∆ r s Let arc AQ = s + ∆s. r A From the right angled trian-

gle △ONP as shown in ﬁg- θ ∆ ure, we have, ←↓ θ → O X Figure 2.5 PN PN sin ∆θ = = ⇒ PN = r sin ∆θ OP r and ON ON cos ∆θ = = ⇒ ON = r cos ∆θ. OP r Also, from the ﬁgure,

NQ = OQ − ON = r + ∆r − r cos ∆θ = r(1 − cos ∆θ) + ∆r ∆θ = 2r sin2 + ∆r. 2 Now from the right angled triangle △PNQ, we have,

PQ2 = PN 2 + NQ2 ∆θ ⇒PQ2 = r2 sin2 ∆θ + (2r sin2 + ∆r)2 [2 ( ) ] ( ) ( ) ( ) ( ) 2 PQ 2 sin ∆θ 2 ∆θ sin ∆θ ∆r ⇒ = r2 + r sin 2 + ∆θ ∆θ 2 ∆θ ∆θ ( ) ( ) ( ) 2 chord PQ 2 arc PQ 2 sin ∆θ 2 ⇒ = r2 arc PQ ∆θ ∆θ [ ( ) ] ( ) ( ) 2 ∆θ sin ∆θ ∆r + r sin 2 + . 2 ∆θ ∆θ 2

Taking Q → P , we get chord PQ → arc PQ. Hence, ( ) ( ) √ ( ) ds 2 dr 2 ds dr 2 = r2 + ⇒ = r2 + . dθ dθ dθ dθ 3. Radius of curvature 5

2.6. Example. For the curve rn = an cos nθ, prove that

ds n−1 = a(sec nθ) n . dθ Solution. Here rn = an cos nθ. Taking log on both the sides, n log r = n log a + log(cos nθ). By diﬀerentiating this we get, n dr sin nθ dr = −n ⇒ r = = −r tan nθ r dθ cos nθ 1 dθ ⇒ 2 2 2 2 2 2 r + r1 = r (1 + tan nθ) = r sec nθ √ ( ) ds dr 2 ⇒ = r2 + dθ dθ

1 n−1 = r sec nθ = a(cos nθ) n sec nθ = a(sec nθ) n .

2.7. Exercise. 1. Show that curvature of a circle is constant and is equal to the reciprocal of its radius. 2. Show that curvature of a straight line is zero. ds 3. Find dx for the following curves. ( ) (i) y = a cosh x . a2 a (ii) y = a log a2−x2 .

ds 4. Find dt for the following curves. (i) x = a(t − sin t); y = a(1 − cos t). (ii) x = a(cos t + t sin t); y = a(sin t − t cos t). (iii) x = aet sin t; y = aet cos t. ds 5. Find dθ for the following curves.

(i) r = a(1 − cos θ). (ii) r2 = a2 cos 2θ.

3. Radius of curvature 3.1. Deﬁnition. Let P be a point on a curve such that the curvature of the curve at P is nonzero. Then the radius of the curvature at P is deﬁned to be the reciprocal of the ds curvature at P and is denoted by ρ. That is, ρ = dψ . 3.2. Theorem. Let y = f(x) be a curve and P be a point on it. Then prove that the radius of curvature at P is given by

3 (1 + y2) 2 ρ = 1 , y2 dy d2y where y1 = dx and y2 = dx2 . 6

Proof. dy Let y = f(x) be the given curve. Then tan ψ = dx . Diﬀerentiating with respect to s, we get, ( ) ( ) dψ d dy d dy dx dx sec2 ψ = = = y ds ds dx dx dx ds 2 ds dψ dx ⇒(1 + tan2 ψ) = y ds 2 ds dψ dx ⇒(1 + y2) = y 1 ds 2 ds ds 1 + y2 ds ⇒ρ = = 1 dψ y dx √ 2 1 + y2 ⇒ 1 2 ρ = 1 + y1 y2 (1 + y2)3/2 ⇒ρ = 1 . y2

3.3. Theorem. Let r = f(θ) be a polar form of a curve with a point P on it. Then prove that the radius of curvature at P is given by (r2 + r2)3/2 ρ = 1 , 2 2 − r + 2r1 rr2 ′ ′′ where r1 = f (θ) and r2 = f (θ). Proof. From the ﬁgure it is clear that ψ = θ + φ. Hence, dψ dθ dφ = + ds ds ds dθ dφ dθ = + ds ( dθ ds ) dθ dφ = 1 + . (3.3.1) φ ds dθ P (r, θ) r We know that tan φ = . Diﬀer- r r1 entiating this with respect to θ, we θ ψ → get, O Figure3.3 2 − 2 dφ r1 rr2 sec φ = 2 dθ r1 dφ r2 − rr 1 r2 − rr 1 r2 − rr ⇒ = 1 2 = 1 2 = 1 2 . dθ r2 1 + tan2 φ r2 r2 r2 + r2 1 1 1 + r2 1 √ 1 We also know that ds = r2 + r2. Hence by (3.3.1), we get, dθ 1 ( ) dψ 1 r2 − rr r2 + 2r2 − rr = √ 1 + 1 2 = 1 2 . ds 2 2 r2 + r2 (r2 + r2)3/2 r + r1 1 1 3. Radius of curvature 7

Hence, ds (r2 + r2)3/2 ρ = = 1 . 2 2 − dψ r + 2r1 rr2

3.4. Example. Prove that if ρ is the radius of curvature at any point P on the parabola y2 = 4ax and S is its focus, then prove that ρ2 ∝ SP 3.

Solution. Let P (x, y) be any point on the give parabola. If the coordinates of the focus S is given by (a, 0), then √ √ SP = (x − a)2 + y2 = x2 − 2ax + a2 + 4ax = x + a.

2 2a Now we ﬁnd ρ for the given parabola y = 4ax. Here 2yy1 = 4a. That is, y1 = y . Also, − 2a − 4a2 y2 = y2 y1 = y3 . Hence,

2 2 3/2 4a 3/2 2 2 3/2 (1 + y ) (1 + 2 ) (y + 4a ) ρ = 1 = y = − y −4a2 4a2 2 y3 (4ax + 4a2)3 64a3(x + a)3 4(x + a)3 4 ⇒ρ2 = = = = SP 3. 16a4 16a4 a a This proves that ρ2 ∝ SP 3.

3.5. Example. Show that the radius of curvature at any point of the curve x = aeθ(cos θ − sin θ), y = aeθ(sin θ + cos θ) is twice the perpendicular distance of the tangent at the point form the origin.

Solution. Here dx = aeθ(cos θ − sin θ) + aeθ(− sin θ − cos θ) = −2aeθ sin θ. dθ Similarly, dy = 2aeθ cos θ. dθ dy − 2 dθ cosec3 θ Hence y1 = dx = cot θ and y2 = cosec θ dx = −2aeθ . Thus, (1 + y2)3/2 (1 + cot2 θ)3/2 ρ = 1 = ( ) = −2aeθ. y − cosec3 θ 2 2aeθ Now the equation of the tangent at a point is dy y − aeθ(sin θ + cos θ) = (x − aeθ(cos θ − sin θ)) dx ⇒y − aeθ(sin θ + cos θ) = − cot θ(x − aeθ(cos θ − sin θ)) ⇒y sin θ + x cos θ − aeθ = 0.

Hence the length of the perpendicular distance of the tangent from the origin is

−aeθ θ − p = cos2 θ+sin2 θ = ae . Hence ρ = 2p. 8

− θ 3.6. Example. For the cycloid x = a(θ + sin θ), y = a(1 cos θ) prove that ρ = 4a cos( 2 ). 2 2 2 Also show that ρ1 + ρ2 = 16a , where ρ1, ρ2 are the radii of curvature at the points where the tangents are perpendicular.

Solution. dx = a(1 + cos θ), dy = a sin θ. Therefore, dθ dθ ( ) ( ) a sin θ 2 sin θ cos θ y = = 2 2 = tan θ 1 a(1 + cos θ) 2 cos2 θ 2 ( )(2 ) 1 dθ 1 1 1 ⇒ y = sec2 θ = = . 2 2 2 dx 2 θ 2 θ 4 θ 2 cos 2 2a cos 2 4a cos 2 Hence, ( ) 2 3/2 3/2 (1 + y1) 2 θ 4 θ 3 θ 4 θ θ ρ = = 1 + tan 4a cos = 4a sec cos = 4a cos( 2 ). y2 2 2 2 2

If P (θ1) and Q(θ2) are the points at which the Y ↑ tangents are perpendicular, θ1 then ρ1 = 4a cos( 2 ) and θ2 ρ2 = 4a cos( 2 ). If the tangents at these points make

the angles ψ1 and ψ2 with the X-axis respectively, then P Q dy θ1 tan ψ1 = dx = tan 2 . ← ψ1 ψ2 → θ1 O X Therefore, ψ1 = 2 . But − π ↓ ψ1 ψ2 = 2 . Therefore, θ1 θ2 π 2 + 2 = 2 . Hence, Figure 3.6 [ ( )] [ ] π θ1 ρ2 + ρ2 = 16a2 cos2 θ1 + cos2 − = 16a2 cos2 θ1 + sin2 θ1 = 16a2. 1 2 2 2 2 2 2

2 3.7. Example. For the curve r = a(1 − cos θ), prove that ρ ∝ r. Also prove that if ρ1 2 2 16a2 and ρ2 are radii of the curvature at the ends of a chord through the pole, ρ1 + ρ2 = 9 .

Solution. Here r1 = a sin θ and r2 = a cos θ. Hence, (r2 + r2)3/2 ρ = 1 2 2 − r + 2r1 rr2 (a2(1 − cos θ)2 + a2 sin2 θ)3/2 = a2(1 − cos θ)2 + 2a2 sin2 θ − a2(1 − cos θ) cos θ (2a2(1 − cos θ))3/2 = 3a2(1 − cos θ) (4a2 sin2 θ )3/2 = 2 2 2 θ 6a sin 2 5. Length of an arc 9

4 = a sin θ . 3 2 Thus, 16 8 8ar ρ2 = a2 sin2 θ = a2(1 − cos θ) = 9 2 9 9 ⇒ ρ2 ∝ r.

Let P (r1, θ1) and P (r2, θ2) be the ends of the chord through the pole. Then θ2 − θ1 = π. Then ρ2 = 16 a2 sin2 θi ,(i = 1, 2). Hence i 9 2 ( ) 16 θ θ ρ2 + ρ2 = a2 sin2 1 + sin2 2 1 2 9 2 2 [ ( )] 16 θ π + θ = a2 sin2 1 + sin2 1 9 2 2 ( ) 16 θ θ = a2 sin2 1 + cos2 1 9 2 2 16 = a2. 9 Rectification 4. Derivative of an arc: Revisited Rectification is the process of computing the length of an arc of a curve. The curves may have different representations – like cartesian, polar and parametric. So, we shall be dealing with all the three forms. Besides, the curve could be expressed as a combination of arcs of two different curves yielding a new closed curve. In this case, the length of arc will be its perimeter. The idea of finding the length of arc is simple. We have obtained the derivative of an arc in Section 2 earlier. It is the derivative of the length of arc s with respect to the independent variable. If we integrate the same, we shall get the length of arc. A curve is said to be rectifiable if it is possible to find its length.

5. Length of an arc of a curve 5.1. Theorem. Let y = f(x) be a cartesian representation of a curve C. Then the length of arc of C between two points A and B corresponding to the x-coordinates a and b respectively, is given by √ ∫b ( ) dy 2 arc AB = 1 + dx. dx a Proof. Let s(x) be the length of arc of curve between ﬁxed point A on the curve and the generic point P (x, f(x)). Then integrating (??) from a to b, we have, √ ∫b ( ) ∫b dy 2 ds 1 + dx = dx dx dx a a 10

∫b = ds

[a ] b = s a = s(b) − s(a) = arc AB − arc AA = arc AB.

5.2. Theorem. Let r = f(θ) be a polar representation of a curve C. Then the length

of arc of C between two points A and B corresponding to the angles θ = θ0 and θ = θ1 respectively, is given by √ ∫θ1 ( ) dr 2 arc AB = r2 + dθ. dθ θ0 5.3. Example. Find the length of arc of the parabola y2 = 4ax,(a > 0), measured from the vertex to one extremity of its latus rectum. Solution. y2 dx y We can write the given equation as x = 4a . Then dy = 2a . Therefore, √ ( ) √ dx 2 y2 1 √ 1 + = 1 + = y2 + 4a2 . dy 4a2 2a ↑

From the ﬁgure, we see that coordinates of the vertex O L(a, 2a) and top end of the latus rec- O → tum L are (0, 0) and (a, 2a) (a, 0) X respectively. Hence the re- L′(a, −2a) quired length of arc is

↓Y √ Figure 5.3 ∫2a ( ) dx 2 arc OL = 1 + dy dy 0 ∫2a 1 √ = y2 + 4a2 dy 2a [0 ] 1 y √ 4a2 √ 2a = y2 + 4a2 + log(y + y2 + 4a2) 2a 2 2 [ 0 ] 1 √ √ = 2 2a2 + 2a2 log(2a + 2a 2) − 0 − 2a2 log 2a 2a 5. Length of an arc 11

[ ( √ )] √ 2a(1 + 2) = a 2 + log 2a (√ √ ) = a 2 + log(1 + 2) .

5.4. Example. (a) Find the entire length of the astroid x2/3 + y2/3 = a2/3. (b) Prove that the length of the curve x2/3 +y2/3 = a2/3 measured from (0, a) to the point 3 2 1/3 (x, y) is given by 2 (ax ) . Solution. (a)

B(0, a)

P (x, y)

O A(a, 0)

Figure 5.4 Here, x2/3 + y2/3 = a2/3 2 2 dy ⇒ x−1/3 + y−1/3 = 0 3 3 dx dy x−1/3 y1/3 ⇒ = − = − dx y−1/3 x1/3 ( ) dy 2 y2/3 a2/3 ⇒1 + = 1 + = = a2/3x−2/3. dx x2/3 x2/3 From the ﬁgure, the entire length of the astroid is √ ∫0 ( ) dy 2 4 × arc AB = 4 1 + dx dx a ∫0 = 4 a1/3x−1/3 dx

a [ ] x2/3 0 = 4a1/3 2/3 a 12 ( ) −a2/3 = 4a1/3 2/3 = −6a.

Since the length of an arc is always positive, we infer that the entire length of the astroid is 6a. (b) The required arc length = arc BP √ ∫x ( ) dy 2 = 1 + dx dx 0 ∫x = a1/3x−1/3 dx 0 [ ] x2/3 x = a1/3 2/3 0 x2/3 = a1/3 2/3 3 = a1/3x2/3 2 3 = (ax2)1/3. 2 √ 5.5. Example. Show that the entire length of the curve x2(a2 − x2) = 8a2y2 is πa 2. Solution. The given curve is symmet-

ric about all – X-axis, Y - ↑ Y axis and the origin. Putting y = 0, we get x ∈ {0, ±a}. x2(a2 − x2) Also since y2 = , √8a2 x a2 − x2 ← →X we have y = √ . (−a, 0)B O A(a, 0) 2 2 a Hence −a ≤ x ≤ a is the only possibility for getting y

real. The shape of the given ↓ curve is as shown in the ﬁg- ure. It contains two equal Figure 5.5 loops. Now,

8a2y2 = x2(a2 − x2) dy ⇒ 16a2y = 2x(a2 − x2) + x2(−2x) dx 5. Length of an arc 13

dy x(a2 − 2x2) ⇒ = dx 8a2y ( ) ( ) dy 2 x(a2 − 2x2) 2 ⇒1 + = 1 + dx 8a2y x2(a2 − 2x2)2 = 1 + 8a2x2(a2 − x2) 8a4 − 8a2x2 + a4 − 4a2x2 + 4x4 = 8a2(a2 − x2) (3a2 − 2x2)2 = . 8a2(a2 − x2) From the ﬁgure (5.5), we say that the entire length = 4 arc OA √ ∫a ( ) dy 2 = 4 1 + dx dx 0

∫a 3a2 − 2x2 = 4 √ √ dx 2a 2 a2 − x2 0 ∫a [ ] a2 a2 − x2 = 4 √ √ + 2 √ √ dx 2a 2 a2 − x2 2a 2 a2 − x2 0 ∫a [ √ ] a a2 − x2 = 4 √ √ + √ dx 2 2 a2 − x2 a 2 0 [ ( ) a x = 4 √ sin−1 2 2 a ( ( ))] 1 x√ a2 x a + √ a2 − x2 + sin−1 a 2 2 2 a √ 0 [ ( ) ]a a x x a2 − x2 = 4 √ sin−1 + √ 2 a 2 2a 0 a π = 4√ 2 √2 = πa 2.

5.6. Example. Find the length of the cardioid r = a(1 + cos θ) lying outside the circle r = −a cos θ. 14

Solution. First we ﬁnd the angle between two curves at the point of their intersection. By comparing them, we get

a(1 + cos θ) = −a cos θ ⇒ θ =

1 π 2 − ⇒ ± ⇒ π cos θ = 2 θ = π 3 3 θ = 2π , 4π . From the ﬁgure, 3 3 A we see that the given curve is symmetric about the po- O B θ = π θ = 0 lar axis and the required arc length is arc ABC = C

4π 2 arc BA. Now for the curve 3 = r = a(1 + cos θ), θ ( ) Figure 5.6 dr 2 r2 + = a2(1 + cos θ)2 + a2 sin2 θ = 2a2(1 + cos θ) = 4a2 cos2 θ . dθ 2 Hence the required arc length √ 2∫π/3 ( ) dr 2 2 arc BA = 2 r2 + dθ dθ 0 2∫π/3 θ = 2 2a cos( 2 ) dθ 0[ ] = 8a sin θ 2π/3 ( 2 0 ) √ π − = 8a sin 3 0 = 4a 3.

6. Intrinsic equation 6.1. Definition. Let A be a fixed point on a curve C and P be a generic point on the curve. Let ψ(P ) denote the angle between the tangents to the curve at points A and P . Also, let s = arc AP . Then the relation between s and ψ is called the intrinsic equation of the curve. It is customary to fix origin (or pole) as the fixed point A if it lies on the curve. Otherwise we mention the fixed point explicitly. We follow this convention throughout this section including exercise. Now we obtain the intrinsic equations of the curve represented in different forms. 6. Intrinsic equation 15

I Cartesian form

↑ Let A(a, b) be a ﬁxed point Y and P (x, y) be a generic point on the curve y = f(x). P We develop the equation in a particular case when the ψ A tangent to the curve at A is ψ → O X parallel to the X-axis. Then √ Figure 6.1 ∫x ( ) dy 2 s = 1 + dx (6.1.1) dx a and dy tan ψ = . (6.1.2) dx Eliminating x from (6.1.1) and (6.1.2) we get a relation F (s, ψ) = 0, which is the intrinsic equation of the curve in cartesian form. If the curve is represented in the form x = f(y) or in a parametric form, then the intrinsic equation can be obtained similarly by eliminating y or the parameter t respectively. However, in the polar form, the coordinates are changed, so we give intrinsic equation in this case separately. II Polar form

Let A(r1, θ1) be a ﬁxed point and P (r, θ) be a generic point on the curve r = f(θ). φ We develop the equation in P a particular case when the ψ tangent to the curve at A A θ ψ → is parallel to the polar axis. O Figure 6.1 Then √ ∫θ ( ) dr 2 s = r2 + dθ. (6.1.3) dθ θ1 Now from the ﬁgure, ψ = θ + φ, (6.1.4) where φ is the angle between the radius vector and the tangent at point P . We also know that dθ tan φ = r . (6.1.5) dr Eliminating φ and θ from (6.1.3), (6.1.4) and (6.1.5) we get F (s, ψ) = 0, (6.1.6) which is the intrinsic equation of the curve in polar form. 16

6.2. Example. Find the intrinsic equation of the Cardioid r = a(1 + cos θ). Hence prove that s2 + 9ρ2 = 16a2, where ρ is the radius of curvature at any point of the curve.

Solution. Here r = a(1 + cos θ). Therefore, dr = −a sin θ. Hence dθ ( ) 2 cos2 θ dθ − 1+cos θ − 2 − θ π θ tan φ = r dr = sin θ = θ θ = cot 2 = tan 2 + 2 . Hence, 2 sin 2 cos 2 π θ φ = + . 2 2 Also, π θ 3θ π ψ = θ + φ = θ + + = + . (6.2.1) 2 2 2 2 Now, √ ∫θ ( ) dr 2 s = r2 + dθ dθ θ1 ∫θ √ = a2 (1 + cos θ)2 + a2 sin2 θ dθ

0 ∫θ √ = a 2 (1 + cos θ) dθ

0 ∫θ √ 2 θ = a 4 cos 2 dθ 0 ∫θ θ = 2a cos 2 dθ [0 ] θ θ = 2a 2 sin 2 0 = 4a sin θ (2 ) ψ π = 4a sin − , (by (6.2.1)) (6.2.2) 3 6 which is the required intrinsic equation. By diﬀerentiating (6.2.2), we have ρ = ds = ( ) ( ) dψ 4a ψ − π ψ − π 2 2 2 3 cos 3 6 . Hence 3ρ = 4a cos 3 6 . So, s + 9ρ = 16a . 6.3. Example. Show that the intrinsic equation of the curve y3 = ax2, is 27s = 8a(sec3 ψ− 1). √ Solution. √1 3/2 dx 3 y We can write the given equation as x = a y . Then dy = 2 a . Here the dx tangent to the curve at the origin is Y -axis. Therefore, tan ψ = dy , That is, √ 3 y tan ψ = . (6.3.1) 2 a 6. Intrinsic equation 17

Now, √ ∫y ( ) dx 2 s = 1 + dy dy ↑Y 0 ∫y √ 9y = 1 + dy 4a 0 [ ] ( ) y P 4a 2 9y 3/2 = 1 + 9 3 4a ψ [( ) 0 ] 9y 3/2 ⇒27s = 8a 1 + − 1 4a [ ] ← → ⇒27s = 8a (1 + tan2 ψ)3/2 − 1 O X Figure 6.3 ⇒27s = 8a(sec3 ψ − 1).

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